Problem Description
problem solving ideas
Except for the step of filling, the rest can be simulated according to the topic, the code is very clear, you can take a look at the code
Code
#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <queue>
#include <cmath>
using namespace std;
int L[8][8];
int M[8][8];
int main()
{
for (int i = 0; i < 8; i ++)
for (int j = 0; j < 8; j ++)
scanf("%d", &L[i][j]); //读入量化矩阵
memset(M, 0, sizeof(M)); //初始化M数组
int n, op;
scanf("%d%d", &n, &op);
int t[64]; //存储输入的n + 1个数
memset(t, 0, sizeof(t));
for (int i = 0; i < n; i ++) scanf("%d", &t[i]);
int i = 0, j = 1; //从（0， 1）开始z字形遍历
int cnt = 0; //记录已经填充了几个数，cnt < n
M[0][0] = t[cnt ++]; //填入（0， 0）这个点
bool up = false; //up标志位记录当前是往右上走还是左下走
while (1)
{
if (!up) //往左下走
{
while (i < 8 && j >= 0 && cnt < n) //i和j没越界并且当前填充个数小于n
{
M[i][j] = t[cnt ++]; //填充
j --; //左
i ++; //下
}
if (cnt == n) break; //已经填充完了，剩下的是0，不用填充了
if (i == 8) //在最下面一行越界了，应该要往右水平走一
{
i --; //i往上一行，为第7行
j += 2; //
if (i == 7 && j == 7) //是最后一格
{
M[i][j] = t[cnt ++]; //填充
break;
}
}
else if (j < 0) j = 0; //左边界越界
up = true; //改方向
continue;
}
else //和上面类似，反一下
{
while (i >= 0 && j < 8 && cnt < n)
{
M[i][j] = t[cnt ++];
i --;
j ++;
}
if (cnt == n) break;
if (j == 8)
{
j --;
i += 2;
}
else if (i < 0) i = 0;
up = false;
continue;
}
}
if (op == 0) //任务0
{
for (int i = 0; i < 8; i ++)
{
for (int j = 0; j < 8; j ++)
{
cout << M[i][j] << " "; //输出填充后的矩阵
}
cout << endl;
}
return 0;
}
if (op == 1) //任务1
{
for (int i = 0; i < 8; i ++)
{
for (int j = 0; j < 8; j ++)
{
M[i][j] *= L[i][j];
cout << M[i][j] << " "; //输出填充矩阵 * 量化矩阵
}
cout << endl;
}
return 0;
}
//任务2
if (op == 2)
{
double sum[8][8]; //记录M‘ij
memset(sum, 0, sizeof(sum));
for (int i = 0; i < 8; i ++)
{
for (int j = 0; j < 8; j ++)
{
double res = 0;
for (int u = 0; u < 8; u ++)
{
for (int v = 0; v < 8; v ++)
{
double t = 1;
if (M[u][v] == 0) continue; //答案为0，不用算了
//照着公式看就行
if (u == 0 && v == 0) t *= 0.5;
else if (u == 0 || v == 0) t *= sqrt(0.5);
t *= M[u][v];
double num1 = 1, num2 = 1;
if (1)
{
num1 *= (i + 0.5);
num1 *= u;
num1 *= acos(-1);
num1 /= 8;
}
if (1)
{
num2 *= (j + 0.5);
num2 *= v;
num2 *= acos(-1);
num2 /= 8;
}
num1 = cos(num1);
num2 = cos(num2);
t *= num1;
t *= num2;
res += t;
}
}
sum[i][j] = res / 4;
}
}
for (int i = 0; i < 8; i ++)
{
for (int j = 0; j < 8; j ++)
{
int x = round(sum[i][j] + 128); //加128取整
if (x < 0) cout << 0 << " ";
else if (x > 255) cout << 255 << " ";
else cout << x << " ";
//cout << sum[i][j] << " ";
}
cout << endl;
}
}
return 0;
}